3 3 Slide Structure of a Waiting Line System n Distribution of Arrivals Generally, the arrival of customers into the system is a random event. Generally, the arrival of customers into the system is a random event. Frequently the arrival pattern is modeled as a Poisson process. Frequently the arrival pattern is modeled as a Poisson process. n Distribution of Service Times Service time is also usually a random variable. Service time is also usually a random variable. A distribution commonly used to describe service time is the exponential distribution. A distribution commonly used to describe service time is the exponential distribution.
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1 Slide Chapter 12 Chapter 12 Waiting Line Models Waiting Line Models The Structure of a Waiting Line System The Structure of a Waiting Line System Queuing Systems Queuing Systems Queuing System Input Characteristics Queuing System Input Characteristics Queuing System Operating Characteristics Queuing System Operating Characteristics Analytical Formulas Analytical Formulas Single-Channel Waiting Line Model with Single-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Poisson Arrivals and Exponential Service Times Times Multiple-Channel Waiting Line Model with Multiple-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Poisson Arrivals and Exponential Service Times Times Economic Analysis of Waiting Lines Economic Analysis of Waiting Lines
Transcript
1 Slide
Chapter 12Chapter 12Waiting Line ModelsWaiting Line Models
The Structure of a Waiting Line SystemThe Structure of a Waiting Line System Queuing SystemsQueuing Systems Queuing System Input CharacteristicsQueuing System Input Characteristics Queuing System Operating CharacteristicsQueuing System Operating Characteristics Analytical FormulasAnalytical Formulas Single-Channel Waiting Line Model with Poisson Single-Channel Waiting Line Model with Poisson
Arrivals and Exponential Service TimesArrivals and Exponential Service Times Multiple-Channel Waiting Line Model with Poisson Multiple-Channel Waiting Line Model with Poisson
Arrivals and Exponential Service TimesArrivals and Exponential Service Times Economic Analysis of Waiting LinesEconomic Analysis of Waiting Lines
2 Slide
Structure of a Waiting Line SystemStructure of a Waiting Line System Queuing theoryQueuing theory is the study of waiting lines. is the study of waiting lines. Four characteristics of a queuing system are: Four characteristics of a queuing system are:
•the manner in which customers arrivethe manner in which customers arrive•the time required for servicethe time required for service•the priority determining the order of servicethe priority determining the order of service•the number and configuration of servers in the number and configuration of servers in
the system.the system.
3 Slide
Structure of a Waiting Line SystemStructure of a Waiting Line System Distribution of ArrivalsDistribution of Arrivals
•Generally, the arrival of customers into the Generally, the arrival of customers into the system is a system is a random eventrandom event. .
•Frequently the arrival pattern is modeled as a Frequently the arrival pattern is modeled as a Poisson processPoisson process..
Distribution of Service TimesDistribution of Service Times•Service time is also usually a random variable. Service time is also usually a random variable.
•A distribution commonly used to describe A distribution commonly used to describe
service time is the service time is the exponential distributionexponential distribution..
4 Slide
Structure of a Waiting Line SystemStructure of a Waiting Line System Queue DisciplineQueue Discipline
•Most common queue discipline is Most common queue discipline is first come, first come, first served (FCFS)first served (FCFS). .
•An elevator is an example of last come, first An elevator is an example of last come, first served (LCFS) queue discipline.served (LCFS) queue discipline.
•Other disciplines assign priorities to the Other disciplines assign priorities to the waiting units and then serve the unit with the waiting units and then serve the unit with the highest priority first.highest priority first.
5 Slide
Structure of a Waiting Line SystemStructure of a Waiting Line System Single Service ChannelSingle Service Channel
Multiple Service ChannelsMultiple Service Channels
SS11
SS11
SS22
SS33
CustomerCustomerleavesleaves
CustomerCustomerleavesleaves
CustomerCustomerarrivesarrives
CustomerCustomerarrivesarrives
Waiting lineWaiting line
Waiting lineWaiting line
SystemSystem
SystemSystem
6 Slide
Queuing SystemsQueuing Systems A A three part codethree part code of the form of the form AA//BB//kk is used to is used to
describe various queuing systems. describe various queuing systems. AA identifies the arrival distribution, identifies the arrival distribution, BB the service the service
(departure) distribution and (departure) distribution and kk the number of the number of channels for the system. channels for the system.
Symbols used for the arrival and service processes Symbols used for the arrival and service processes are: are: MM - Markov distributions - Markov distributions (Poisson/exponential), (Poisson/exponential), DD - Deterministic (constant) - Deterministic (constant) and and GG - General distribution (with a known mean - General distribution (with a known mean and variance). and variance).
For example, For example, MM//MM//kk refers to a system in which refers to a system in which arrivals occur according to a Poisson distribution, arrivals occur according to a Poisson distribution, service times follow an exponential distribution service times follow an exponential distribution and there are and there are kk servers working at identical servers working at identical service rates. service rates.
7 Slide
Queuing System Input CharacteristicsQueuing System Input Characteristics
= the average arrival = the average arrival raterate 1/1/ = the average = the average timetime between arrivals between arrivals µ µ = the average service = the average service raterate for each server for each server 1/1/µ µ = the average service = the average service timetime = the standard deviation of the service = the standard deviation of the service timetime
8 Slide
Queuing System Operating CharacteristicsQueuing System Operating Characteristics
PP0 0 = probability the service facility is idle = probability the service facility is idlePPnn = probability of = probability of nn units in the system units in the systemPPww = probability an arriving unit must wait for = probability an arriving unit must wait for
serviceservice LLqq = average number of units in the queue = average number of units in the queue
awaiting awaiting serviceservice LL = average number of units in the system = average number of units in the systemWWqq = average time a unit spends in the queue = average time a unit spends in the queue
awaiting serviceawaiting service WW = average time a unit spends in the system = average time a unit spends in the system
9 Slide
Analytical FormulasAnalytical Formulas For nearly all queuing systems, there is a For nearly all queuing systems, there is a
relationship between the average time a unit relationship between the average time a unit spends in the system or queue and the spends in the system or queue and the average number of units in the system or average number of units in the system or queue. queue.
These relationships, known as These relationships, known as Little's flow Little's flow equationsequations are: are:
LL = = WW and and LLqq = = WWqq
10 Slide
Analytical FormulasAnalytical Formulas When the queue discipline is FCFS, analytical When the queue discipline is FCFS, analytical
formulas have been derived for several different formulas have been derived for several different queuing models including the following: queuing models including the following: •MM//MM/1/1•MM//MM//kk•MM//GG/1/1•MM//GG//kk with blocked customers cleared with blocked customers cleared•MM//MM/1 with a finite calling population/1 with a finite calling population
Analytical formulas are not available for all Analytical formulas are not available for all possible queuing systems. In this event, possible queuing systems. In this event, insights may be gained through a simulation of insights may be gained through a simulation of the system. the system.
11 Slide
M/M/1 Queuing SystemM/M/1 Queuing System Single channelSingle channel Poisson arrival-rate distributionPoisson arrival-rate distribution Exponential service-time distributionExponential service-time distribution Unlimited maximum queue lengthUnlimited maximum queue length Infinite calling populationInfinite calling population Examples:Examples:
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) MM//MM/1 Queuing System/1 Queuing System
Joe Ferris is a stock trader on the floor of Joe Ferris is a stock trader on the floor of the New York Stock Exchange for the firm of the New York Stock Exchange for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per transactions arrive at a mean rate of 20 per hour. Each order received by Joe requires an hour. Each order received by Joe requires an average of two minutes to process.average of two minutes to process.
Orders arrive at a mean rate of 20 per Orders arrive at a mean rate of 20 per hour or one order every 3 minutes. Therefore, hour or one order every 3 minutes. Therefore, in a 15 minute interval the average number of in a 15 minute interval the average number of orders arriving will be orders arriving will be = 15/3 = 5. = 15/3 = 5.
13 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Arrival Rate DistributionArrival Rate DistributionQuestionQuestion
What is the probability that no orders What is the probability that no orders are received within a 15-minute period?are received within a 15-minute period?AnswerAnswer
P P ((xx = 0) = (5 = 0) = (500e e -5-5)/0! = )/0! = e e -5-5 = = .0067.0067
14 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Arrival Rate DistributionArrival Rate DistributionQuestionQuestion
What is the probability that exactly 3 What is the probability that exactly 3 orders are received within a 15-minute period?orders are received within a 15-minute period?AnswerAnswer
P P ((xx = 3) = (5 = 3) = (533e e -5-5)/3! = 125(.0067)/6 = )/3! = 125(.0067)/6 = .1396.1396
15 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Arrival Rate DistributionArrival Rate DistributionQuestionQuestionWhat is the probability that more than 6 orders What is the probability that more than 6 orders arrive within a 15-minute period? arrive within a 15-minute period? AnswerAnswer P P ((xx > 6) = 1 - > 6) = 1 - P P ((xx = 0) - = 0) - P P ((xx = 1) - = 1) - P P ((xx = = 2) 2) - - P P ((xx = 3) - = 3) - P P ((xx = 4) - = 4) - P P ((xx = 5) = 5)
- - P P ((xx = 6) = 6) = 1 - .762 = = 1 - .762 = .238.238
16 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Service Rate DistributionService Rate DistributionQuestionQuestion
What is the mean service rate per hour?What is the mean service rate per hour?AnswerAnswer
Since Joe Ferris can process an order in an Since Joe Ferris can process an order in an average time of 2 minutes (= 2/60 hr.), then the average time of 2 minutes (= 2/60 hr.), then the mean service rate, mean service rate, µµ, is , is µµ = 1/(mean service = 1/(mean service time), or 60/2.time), or 60/2.
= 30/hr.= 30/hr.
17 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Service Time DistributionService Time DistributionQuestionQuestionWhat percentage of the orders will take less than What percentage of the orders will take less than one minute to process? one minute to process? AnswerAnswerSince the units are expressed in hours, Since the units are expressed in hours, P P ((TT << 1 minute) = 1 minute) = P P ((TT << 1/60 hour). 1/60 hour). Using the exponential distribution, Using the exponential distribution, P P ((TT << t t ) = 1 - ) = 1 - ee-µt-µt. . Hence, Hence, P P ((TT << 1/60) = 1 - e 1/60) = 1 - e-30(1/60)-30(1/60) = 1 - .6065 = .3935 = = 1 - .6065 = .3935 = 39.35%39.35%
18 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Service Time DistributionService Time DistributionQuestionQuestion
What percentage of the orders will be What percentage of the orders will be processed in exactly 3 minutes?processed in exactly 3 minutes?AnswerAnswer
Since the exponential distribution is a Since the exponential distribution is a continuous distribution, the probability a continuous distribution, the probability a service time exactly equals any specific value is service time exactly equals any specific value is 0 .0 .
19 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Service Time DistributionService Time DistributionQuestionQuestion
What percentage of the orders will require What percentage of the orders will require more than 3 minutes to process?more than 3 minutes to process?AnswerAnswer
The percentage of orders requiring more The percentage of orders requiring more than 3 minutes to process is:than 3 minutes to process is: P P ((TT > 3/60) = > 3/60) = ee-30(3/60)-30(3/60) = = ee -1.5-1.5 = .2231 = = .2231 = 22.31% 22.31%
20 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Average Time in the SystemAverage Time in the SystemQuestionQuestionWhat is the average time an order must wait from What is the average time an order must wait from the time Joe receives the order until it is finished the time Joe receives the order until it is finished being processed (i.e. its turnaround time)?being processed (i.e. its turnaround time)?AnswerAnswerThis is an This is an MM//MM/1 queue with /1 queue with = 20 per hour and = 20 per hour and = 30 per hour. The average time an order waits = 30 per hour. The average time an order waits in the system is:in the system is: WW = 1/(µ - = 1/(µ - ) ) = 1/(30 - 20)= 1/(30 - 20) = 1/10 hour or = 1/10 hour or 6 minutes6 minutes
21 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Average Length of QueueAverage Length of QueueQuestionQuestionWhat is the average number of orders Joe has What is the average number of orders Joe has waiting to be processed?waiting to be processed?AnswerAnswerAverage number of orders waiting in the queue is:Average number of orders waiting in the queue is: LLqq = = 22/[µ(µ - /[µ(µ - )] )] = (20)= (20)22/[(30)(30-20)]/[(30)(30-20)] = 400/300 = 400/300 = = 4/34/3
22 Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A) Utilization FactorUtilization FactorQuestionQuestion
What percentage of the time is Joe What percentage of the time is Joe processing orders?processing orders?AnswerAnswer
The percentage of time Joe is processing The percentage of time Joe is processing orders is equivalent to the utilization factor, orders is equivalent to the utilization factor, //. . Thus, the percentage of time he is processing Thus, the percentage of time he is processing orders is:orders is:
